In dimensional analysis, we might consider a problem like:

$$ f(q_1, q_2, ..., q_{n-1}) = q_{n} $$

where $q_i$ are physical quantities of interest. In order to determine how many $\Pi$ groups can be formed, we consider how many "fundamental" quantities are present in the problem. Let us say there are $k$ such fundamental quantities, then we will have $j = n - k$ dimensionless groups, and we can write the above relationship as:

$$ f(\Pi_1, \Pi_2, ..., \Pi_{n-k-1}) = \Pi_{n-k}$$

So consider this in the case of an example, the equation of motion in a 1-D overdamped system, with $\eta$ being the viscosity (dimensions $MT^{-1}$), velocity $\rm{d}x/\rm{d}t$ with units $LT^{-1}$, and force $F$ with units $MLT^{-2}$:

$$\eta \frac{\rm{d}x}{\rm{d}t} = F$$

Or, I can rewrite the equation in a form more familiar in the context of dimensional analysis, where don't care about the actual relationship between quantities: $$ f(\eta, \frac{\rm{d}x}{\rm{d}t}, F) = 0$$

Here, we have $n = 3$ variables, and one may say that we also have $k = 3$ fundamental dimensions $M$, $L$ and $T$. Then, we have $j = 0$! What now?

$$ f(1, 1, 1) = 0$$

What is dimensional analysis telling me here? Is dimensional analysis telling me that I can write the relationship between variables in a totally dimensionless form?

Okay, maybe I can dance around this by saying that I really only have $k = 2$ fundamental dimensions, if I let them be $MT^{-1} = [\eta]$, and $LT^{-1} = [\dot{x}]$ respectively. Okay, so now I have $j = 1$, so I can form one $\Pi$ group, let me call it $\Pi_0$. Skipping the work, $\Pi_0$ will correspond to $F = \eta\dot{x}$.

This is a major hint that dimensional analysis is telling me something really interesting. In cases where we have $j \leq 0$, we can redefine "fundamental dimensions" to have $j > 0$. Then, dimensional analysis gives us a...major hint about how the variables are related? I am sure there is a better (more insightful) way to put what I have I just learned -- but what is it?

In dimensional analysis, we might consider a problem like:

$$ f(q_1, q_2, ..., q_{n-1}) = q_{n} $$

where $q_i$ are physical quantities of interest. In order to determine how many $\Pi$ groups can be formed, we consider how many "fundamental" quantities are present in the problem. Let us say there are $k$ such fundamental quantities, then we will have $j = n - k$ dimensionless groups, and we can write the above relationship as:

$$ f(\Pi_1, \Pi_2, ..., \Pi_{n-k-1}) = \Pi_{n-k}$$

So consider this in the case of an example, the equation of motion in a 1-D overdamped system, with $\eta$ being the viscosity (dimensions $MT^{-1}$), velocity $\rm{d}x/\rm{d}t$ with units $LT^{-1}$, and force $F$ with units $MLT^{-2}$:

$$\eta \frac{\rm{d}x}{\rm{d}t} = F$$

Or, I can rewrite the equation in a form more familiar in the context of dimensional analysis, where don't care about the actual relationship between quantities: $$ f(\eta, \frac{\rm{d}x}{\rm{d}t}, F) = 0$$

Here, we have $n = 3$ variables, and one may say that we also have $k = 3$ fundamental dimensions $M$, $L$ and $T$. Then, we have $j = 0$! What now?

$$ f(1, 1, 1) = 0$$

What is dimensional analysis telling me here? Is dimensional analysis telling me that I can write the relationship between variables in a totally dimensionless form?

Okay, maybe I can dance around this by saying that I really only have $k = 2$ fundamental dimensions, if I let them be $MT^{-1} = [\eta]$, and $LT^{-1} = [\dot{x}]$. Okay, so now I have $j = 1$, so I can form one $\Pi$ group, let me call it $\Pi_0$. Skipping the work, $\Pi_0$ will correspond to $F = \eta\dot{x}$.

This is a major hint that dimensional analysis is telling me something really interesting. In cases where we have $j \leq 0$, we can redefine "fundamental dimensions" to have $j > 0$. Then, dimensional analysis gives us a...major hint about how the variables are related? I am sure there is a better (more insightful) way to put what I have I just learned -- but what is it?